$ E = \left[\begin{array}{rr}5 & 3 \\ -2 & 1 \\ 0 & -2\end{array}\right]$ $ B = \left[\begin{array}{rr}4 & 2 \\ 5 & -2\end{array}\right]$ What is $ E B$ ?
Solution: Because $ E$ has dimensions $(3\times2)$ and $ B$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(3\times2)$ $ E B = \left[\begin{array}{rr}{5} & {3} \\ {-2} & {1} \\ \color{gray}{0} & \color{gray}{-2}\end{array}\right] \left[\begin{array}{rr}{4} & \color{#DF0030}{2} \\ {5} & \color{#DF0030}{-2}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ E$ , with the corresponding elements in column $j$ of the second matrix, $ B$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ E$ with the first element in ${\text{column }1}$ of $ B$ , then multiply the second element in ${\text{row }1}$ of $ E$ with the second element in ${\text{column }1}$ of $ B$ , and so on. Add the products together. $ \left[\begin{array}{rr}{5}\cdot{4}+{3}\cdot{5} & ? \\ ? & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ E$ with the corresponding elements in ${\text{column }1}$ of $ B$ and add the products together. $ \left[\begin{array}{rr}{5}\cdot{4}+{3}\cdot{5} & ? \\ {-2}\cdot{4}+{1}\cdot{5} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ E$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ B$ and add the products together. $ \left[\begin{array}{rr}{5}\cdot{4}+{3}\cdot{5} & {5}\cdot\color{#DF0030}{2}+{3}\cdot\color{#DF0030}{-2} \\ {-2}\cdot{4}+{1}\cdot{5} & ? \\ ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{5}\cdot{4}+{3}\cdot{5} & {5}\cdot\color{#DF0030}{2}+{3}\cdot\color{#DF0030}{-2} \\ {-2}\cdot{4}+{1}\cdot{5} & {-2}\cdot\color{#DF0030}{2}+{1}\cdot\color{#DF0030}{-2} \\ \color{gray}{0}\cdot{4}+\color{gray}{-2}\cdot{5} & \color{gray}{0}\cdot\color{#DF0030}{2}+\color{gray}{-2}\cdot\color{#DF0030}{-2}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}35 & 4 \\ -3 & -6 \\ -10 & 4\end{array}\right] $